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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: the length of the arc Length of Arc APB = 𝜽/𝟑𝟔𝟎 × (𝟐𝝅𝒓) = (60°)/(360°) × 2 × 22/7 × 21 = 1/6 × 2 × 22 × 3 = 22 cm Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (ii) area of the sector formed by the arc Area of sector OAPB = 𝜃/360×𝜋𝑟2 = 𝟔𝟎/𝟑𝟔𝟎 × 𝟐𝟐/𝟕 × 𝟐𝟏 × 𝟐𝟏 = 1/6 × 22/7 × 21 × 21 = 1/6 × 22 × 3 × 21 = 231 cm2 Ex 11.1, 5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (iii) area of segment formed by the corresponding chord Area of segment APB = Area of sector OAPB – Area of ΔOAB From last part, Area of sector OAPB = 231 cm2 Finding area of Δ AOB Area Δ AOB = 1/2 × Base × Height We draw OM ⊥ AB ∴ ∠ OMB = ∠ OMA = 90° And, by symmetry M is the mid-point of AB ∴ BM = AM = 1/2 AB In right triangle Δ OMA sin O = (side opposite to angle O)/Hypotenuse sin 𝟑𝟎° = 𝐀𝑴/𝑨𝑶 1/2=𝐴𝑀/21 21/2 = AM AM = 𝟐𝟏/𝟐 In right triangle Δ OMA cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝟑𝟎° = 𝑶𝑴/𝑨𝑶 √3/2=𝑂𝑀/21 √3/2 × 21 = OM OM = √𝟑/𝟐 × 21 From (1) AM = 𝟏/𝟐AB 2AM = AB AB = 2AM Putting value of AM AB = 2 × 1/2 × 21 AB = 21cm Now, Area of Δ AOB = 1/2 × Base × Height = 𝟏/𝟐 × AB × OM = 1/2 × 21 × √3/2 × 21 = (𝟒𝟒𝟏√𝟑)/𝟒 cm2 Therefore, Area of segment APB = Area of sector OAPB – Area of ΔOAB = (231 – 𝟒𝟒𝟏/𝟒 √𝟑) cm2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.